3. A framed photograph has a uniform width matte around the photo. The outside of the frame has dimensions of $25 \mathrm{~cm}$ by $30 \mathrm{~cm}$. If the area of the matte is twice the area of the photo, what is the width of the matte? [T-5 marks]

Step 1 ：The area of the frame is \(25\,\text{cm} \times 30\,\text{cm} = 750\,\text{cm}^2\).

Step 2 ：Let's denote the width of the matte as \(x\). The photo then has dimensions of \((25-2x)\,\text{cm}\) by \((30-2x)\,\text{cm}\).

Step 3 ：According to the problem, the area of the matte is twice the area of the photo. This gives us the equation: \(750\,\text{cm}^2 - (25-2x)(30-2x) = 2*(25-2x)(30-2x)\).

Step 4 ：We can solve this equation to find the value of \(x\).

Step 5 ：The solutions to the equation are \(\frac{55}{4} - \frac{5\sqrt{41}}{4}\) and \(\frac{5\sqrt{41}}{4} + \frac{55}{4}\).

Step 6 ：However, the width of the matte cannot be negative, so we discard the negative solution.

Step 7 ：Therefore, the width of the matte is \(\frac{5\sqrt{41}}{4} + \frac{55}{4}\).

Step 8 ：Final Answer: The width of the matte is \(\boxed{\frac{5\sqrt{41}}{4} + \frac{55}{4}}\) cm.